ACM--模拟--HDOJ 1008--Elevator

Elevator

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 62680    Accepted Submission(s): 34449

Problem Description

The highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator will stop, in specified order. It costs 6 seconds to move the elevator up one floor, and 4 seconds to move down one floor. The elevator will stay for 5 seconds at each stop.

For a given request list, you are to compute the total time spent to fulfill the requests on the list. The elevator is on the 0th floor at the beginning and does not have to return to the ground floor when the requests are fulfilled.

 

Input

There are multiple test cases. Each case contains a positive integer N, followed by N positive numbers. All the numbers in the input are less than 100. A test case with N = 0 denotes the end of input. This test case is not to be processed.

 

Output

Print the total time on a single line for each test case. 

 

Sample Input

1 2 3 2 3 1 0

 

分析：这里存在一只特殊情况，两个数字相同，即待在一层楼两次，只需要加5即可

#include<iostream>
using namespace std;
#include<cmath>
int a[120];
int main()
{
	int n,sum;
	while(cin>>n&&n!=0){
		sum=0;
		for(int i=1;i<=n;i++){ 
			cin>>a[i];
			if(i==1){
				if(a[i]!=0)
					sum+=6*a[i]+5;
				else    ；可有可无
					sum+=0;
			}
			else if(a[i]>a[i-1])
				sum+=6*(a[i]-a[i-1])+5;
			else if(a[i]<a[i-1])
				sum+=4*(a[i-1]-a[i])+5;	
			else if(a[i]==a[i-1])    ；一开始没有考虑这个
				sum+=5;	 
		} 
		printf("%d\n",sum);
	}
	return 0;
}

Sample Output

17 41